geometry parabola problem

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#1 h Apr 11, 2025, 9:52 PM • 1 Y

Y by Kizaruno

How would you solve this without using calculus?

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ShadowDragonRules

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#2 h Apr 11, 2025, 10:12 PM • 1 Y

Y by Kizaruno

hmm... I would suggest to find a relationship of graphing this through first finding the parabola's graph to find the eventual diameter. BTW

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#3 h Apr 11, 2025, 11:15 PM • 1 Y

Y by Kizaruno

smalkaram_3549 wrote:

How would you solve this without using calculus?

is it $R=\frac{\sqrt{11}}{2}$ ?
I found it solving a quadratic equation.

This post has been edited 1 time. Last edited by mathmax001, Apr 11, 2025, 11:16 PM
Reason: mistype

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#4 h Apr 11, 2025, 11:42 PM

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Solution 1:

$y'=2x$ , so the normal line through $(a,a^2)$ is $y=a^2-\frac{x-a}{2a}$ ,

Put $(x,y)=(0,3)$ to get $a^2+\frac{1}{2}=3\implies a^2=\frac{5}{2}$ .

So the points of tangency are $\left(\pm\frac{\sqrt{10}}{2},\frac{5}{2}\right)$ .

$r=\sqrt{\left(\frac{\sqrt{10}}{2}-0\right)^2+\left(\frac{5}{2}-3\right)^2}=\sqrt{\frac{5}{2}+\frac{1}{4}}=\frac{\sqrt{11}}{2}$ .

Solution 2:

$r(t)=\text{dist}[(0,3),(t,t^2)]=\sqrt{t^2+(t^2-3)^2}=\sqrt{t^4-5t^2+9}=\sqrt{\left(t^2-\frac{5}{2}\right)^2+\frac{11}{4}}\geq\frac{\sqrt{11}}{2}$ with equality when $t^2=\frac{5}{2}$ .

This post has been edited 1 time. Last edited by rchokler, Apr 11, 2025, 11:43 PM

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#5 h Apr 12, 2025, 12:05 AM • 1 Y

Y by Kizaruno

mathmax001 wrote:

smalkaram_3549 wrote:

How would you solve this without using calculus?

is it $R=\frac{\sqrt{11}}{2}$ ?
I found it solving a quadratic equation.

yes it is. How did you do that? I got it using calculus but can't figure out the algebraic method

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#6 h Apr 12, 2025, 12:25 AM

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We let the circle equation be $x^2+(y-3)^2=r^2$ . We can solve for $y$ :
$y^2-6y+9+y=r^2.$ Due to tangency, there is only one solution of $y$ , and we have a double root and
$5^2=4\left(9-r^2\right)\implies r=\boxed{\frac{\sqrt{11}}2}.$ @rchokler's solution 2 is also algebraic and works well.

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ReticulatedPython

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#7 h Apr 12, 2025, 12:33 AM

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Solution

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#8 h Apr 12, 2025, 7:30 PM

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ReticulatedPython wrote:

Solution

This is exactly the method I used .

If you don't understand the meaning of this sentence " Since we want the roots to be negations of each other, the discriminant is equal to $0.$ " , I'll explain it more :
If the discriminant $5^2-4(9-r^2)$ were $\neq 0$ then the equation would have 2 different solutions for $x^2$ ( because the discriminant is positive ) , that means four 4 points of tangency which is not true .

This post has been edited 2 times. Last edited by mathmax001, Apr 12, 2025, 7:41 PM
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#9 h Apr 12, 2025, 7:41 PM • 2 Y

Y by Kizaruno, ReticulatedPython

sol

also hi reticulated python!

also sorry for a bad sol
i didnt have time to explain everything clearly

This post has been edited 1 time. Last edited by jb2015007, Apr 12, 2025, 7:42 PM

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#11 h Apr 13, 2025, 3:40 AM

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The equation of the circle is $x^2 + (y - 3)^2 = r^2$ . Subbing in $y = x^2$ to find the intersection gets $x^2 + (x^2 - 3)^2 = r^2$ which is $x^4 - 5x^2 + 9 - r^2 = 0$ now since, the circle only intersects the parabola twice there must be 2 double roots so we consider the determinant which is $5^2 - 4(9 - r^2) = 0$ which becomes $4r^2 - 11 = 0$ so $r = \sqrt{11}/2$ only since the radius can't be negative.

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ReticulatedPython

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#12 h Apr 13, 2025, 2:05 PM

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jb2015007 wrote:

sol

also hi reticulated python!

also sorry for a bad sol
i didnt have time to explain everything clearly

Looks good to me!

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